Integrand size = 15, antiderivative size = 302 \[ \int (e+f x)^3 \text {arctanh}(\tan (a+b x)) \, dx=\frac {i (e+f x)^4 \arctan \left (e^{2 i (a+b x)}\right )}{4 f}+\frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 f^3 \operatorname {PolyLog}\left (5,-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 f^3 \operatorname {PolyLog}\left (5,i e^{2 i (a+b x)}\right )}{16 b^4} \]
1/4*I*(f*x+e)^4*arctan(exp(2*I*(b*x+a)))/f+1/4*(f*x+e)^4*arctanh(tan(b*x+a ))/f-1/4*I*(f*x+e)^3*polylog(2,-I*exp(2*I*(b*x+a)))/b+1/4*I*(f*x+e)^3*poly log(2,I*exp(2*I*(b*x+a)))/b+3/8*f*(f*x+e)^2*polylog(3,-I*exp(2*I*(b*x+a))) /b^2-3/8*f*(f*x+e)^2*polylog(3,I*exp(2*I*(b*x+a)))/b^2+3/8*I*f^2*(f*x+e)*p olylog(4,-I*exp(2*I*(b*x+a)))/b^3-3/8*I*f^2*(f*x+e)*polylog(4,I*exp(2*I*(b *x+a)))/b^3-3/16*f^3*polylog(5,-I*exp(2*I*(b*x+a)))/b^4+3/16*f^3*polylog(5 ,I*exp(2*I*(b*x+a)))/b^4
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(654\) vs. \(2(302)=604\).
Time = 1.06 (sec) , antiderivative size = 654, normalized size of antiderivative = 2.17 \[ \int (e+f x)^3 \text {arctanh}(\tan (a+b x)) \, dx=\frac {1}{4} x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right ) \text {arctanh}(\tan (a+b x))+\frac {-8 b^4 e^3 x \log \left (1-i e^{2 i (a+b x)}\right )-12 b^4 e^2 f x^2 \log \left (1-i e^{2 i (a+b x)}\right )-8 b^4 e f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )-2 b^4 f^3 x^4 \log \left (1-i e^{2 i (a+b x)}\right )+8 b^4 e^3 x \log \left (1+i e^{2 i (a+b x)}\right )+12 b^4 e^2 f x^2 \log \left (1+i e^{2 i (a+b x)}\right )+8 b^4 e f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )+2 b^4 f^3 x^4 \log \left (1+i e^{2 i (a+b x)}\right )-4 i b^3 (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+4 i b^3 (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )+6 b^2 e^2 f \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )+12 b^2 e f^2 x \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )+6 b^2 f^3 x^2 \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b^2 e^2 f \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )-12 b^2 e f^2 x \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )-6 b^2 f^3 x^2 \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )+6 i b e f^2 \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )+6 i b f^3 x \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-6 i b e f^2 \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )-6 i b f^3 x \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )-3 f^3 \operatorname {PolyLog}\left (5,-i e^{2 i (a+b x)}\right )+3 f^3 \operatorname {PolyLog}\left (5,i e^{2 i (a+b x)}\right )}{16 b^4} \]
(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)*ArcTanh[Tan[a + b*x]])/4 + (-8*b^4*e^3*x*Log[1 - I*E^((2*I)*(a + b*x))] - 12*b^4*e^2*f*x^2*Log[1 - I* E^((2*I)*(a + b*x))] - 8*b^4*e*f^2*x^3*Log[1 - I*E^((2*I)*(a + b*x))] - 2* b^4*f^3*x^4*Log[1 - I*E^((2*I)*(a + b*x))] + 8*b^4*e^3*x*Log[1 + I*E^((2*I )*(a + b*x))] + 12*b^4*e^2*f*x^2*Log[1 + I*E^((2*I)*(a + b*x))] + 8*b^4*e* f^2*x^3*Log[1 + I*E^((2*I)*(a + b*x))] + 2*b^4*f^3*x^4*Log[1 + I*E^((2*I)* (a + b*x))] - (4*I)*b^3*(e + f*x)^3*PolyLog[2, (-I)*E^((2*I)*(a + b*x))] + (4*I)*b^3*(e + f*x)^3*PolyLog[2, I*E^((2*I)*(a + b*x))] + 6*b^2*e^2*f*Pol yLog[3, (-I)*E^((2*I)*(a + b*x))] + 12*b^2*e*f^2*x*PolyLog[3, (-I)*E^((2*I )*(a + b*x))] + 6*b^2*f^3*x^2*PolyLog[3, (-I)*E^((2*I)*(a + b*x))] - 6*b^2 *e^2*f*PolyLog[3, I*E^((2*I)*(a + b*x))] - 12*b^2*e*f^2*x*PolyLog[3, I*E^( (2*I)*(a + b*x))] - 6*b^2*f^3*x^2*PolyLog[3, I*E^((2*I)*(a + b*x))] + (6*I )*b*e*f^2*PolyLog[4, (-I)*E^((2*I)*(a + b*x))] + (6*I)*b*f^3*x*PolyLog[4, (-I)*E^((2*I)*(a + b*x))] - (6*I)*b*e*f^2*PolyLog[4, I*E^((2*I)*(a + b*x)) ] - (6*I)*b*f^3*x*PolyLog[4, I*E^((2*I)*(a + b*x))] - 3*f^3*PolyLog[5, (-I )*E^((2*I)*(a + b*x))] + 3*f^3*PolyLog[5, I*E^((2*I)*(a + b*x))])/(16*b^4)
Time = 1.18 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6805, 3042, 4669, 3011, 7163, 7163, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int (e+f x)^3 \text {arctanh}(\tan (a+b x)) \, dx\) |
| \(\Big \downarrow \) 6805 |
| \(\displaystyle \frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {b \int (e+f x)^4 \sec (2 a+2 b x)dx}{4 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {b \int (e+f x)^4 \csc \left (2 a+2 b x+\frac {\pi }{2}\right )dx}{4 f}\) |
| \(\Big \downarrow \) 4669 |
| \(\displaystyle \frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {b \left (-\frac {2 f \int (e+f x)^3 \log \left (1-i e^{2 i (a+b x)}\right )dx}{b}+\frac {2 f \int (e+f x)^3 \log \left (1+i e^{2 i (a+b x)}\right )dx}{b}-\frac {i (e+f x)^4 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{4 f}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {b \left (\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \int (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \int (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {i (e+f x)^4 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{4 f}\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {b \left (\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \left (\frac {i f \int (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )dx}{b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{2 b}\right )}{b}-\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \left (\frac {i f \int (e+f x) \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )dx}{b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^4 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{4 f}\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {b \left (\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \left (\frac {i f \left (\frac {i f \int \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (e+f x) \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{2 b}\right )}{b}-\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \left (\frac {i f \left (\frac {i f \int \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (e+f x) \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^4 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{4 f}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {b \left (\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \left (\frac {i f \left (\frac {f \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{2 b}\right )}{b}-\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \left (\frac {i f \left (\frac {f \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^4 \arctan \left (e^{2 i (a+b x)}\right )}{b}\right )}{4 f}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {(e+f x)^4 \text {arctanh}(\tan (a+b x))}{4 f}-\frac {b \left (-\frac {i (e+f x)^4 \arctan \left (e^{2 i (a+b x)}\right )}{b}+\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \left (\frac {i f \left (\frac {f \operatorname {PolyLog}\left (5,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{2 b}\right )}{2 b}\right )}{b}-\frac {2 f \left (\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{2 b}-\frac {3 i f \left (\frac {i f \left (\frac {f \operatorname {PolyLog}\left (5,i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{2 b}\right )}{b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{2 b}\right )}{2 b}\right )}{b}\right )}{4 f}\) |
((e + f*x)^4*ArcTanh[Tan[a + b*x]])/(4*f) - (b*(((-I)*(e + f*x)^4*ArcTan[E ^((2*I)*(a + b*x))])/b + (2*f*(((I/2)*(e + f*x)^3*PolyLog[2, (-I)*E^((2*I) *(a + b*x))])/b - (((3*I)/2)*f*(((-1/2*I)*(e + f*x)^2*PolyLog[3, (-I)*E^(( 2*I)*(a + b*x))])/b + (I*f*(((-1/2*I)*(e + f*x)*PolyLog[4, (-I)*E^((2*I)*( a + b*x))])/b + (f*PolyLog[5, (-I)*E^((2*I)*(a + b*x))])/(4*b^2)))/b))/b)) /b - (2*f*(((I/2)*(e + f*x)^3*PolyLog[2, I*E^((2*I)*(a + b*x))])/b - (((3* I)/2)*f*(((-1/2*I)*(e + f*x)^2*PolyLog[3, I*E^((2*I)*(a + b*x))])/b + (I*f *(((-1/2*I)*(e + f*x)*PolyLog[4, I*E^((2*I)*(a + b*x))])/b + (f*PolyLog[5, I*E^((2*I)*(a + b*x))])/(4*b^2)))/b))/b))/b))/(4*f)
3.4.12.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol ] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si mp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x ))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Int[ArcTanh[Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[Tan[a + b*x]]/(f*(m + 1))), x] - Simp[b/ (f*(m + 1)) Int[(e + f*x)^(m + 1)*Sec[2*a + 2*b*x], x], x] /; FreeQ[{a, b , e, f}, x] && IGtQ[m, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. )*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F])) Int[(e + f*x) ^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c , d, e, f, n, p}, x] && GtQ[m, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 13.45 (sec) , antiderivative size = 3640, normalized size of antiderivative = 12.05
1/2/b*e^3*a*ln(exp(2*I*(b*x+a))+I)-3/4*f*e^2*ln(-I*exp(2*I*(b*x+a))+1)*x^2 -3/8*f^3/b^2*polylog(3,I*exp(2*I*(b*x+a)))*x^2-3/8*f/b^2*e^2*polylog(3,I*e xp(2*I*(b*x+a)))+3/8*f/b^2*e^2*polylog(3,-I*exp(2*I*(b*x+a)))+3/8*f^3/b^2* polylog(3,-I*exp(2*I*(b*x+a)))*x^2-3/16*f^3*polylog(5,-I*exp(2*I*(b*x+a))) /b^4+3/16*f^3*polylog(5,I*exp(2*I*(b*x+a)))/b^4-1/2*f^3/b^3*a^3*ln(1+exp(I *(b*x+a))*(-1)^(3/4))*x-1/2*f^3/b^3*a^3*ln(1-exp(I*(b*x+a))*(-1)^(3/4))*x- f^2/b^3*e*ln(1+I*exp(2*I*(b*x+a)))*a^3+1/2*f^3/b^3*ln(1+I*exp(2*I*(b*x+a)) )*x*a^3-1/2*f^2/b^3*a^3*e*ln(-exp(2*I*(b*x+a))+I)+3/4*f/b^2*a^2*e^2*ln(-ex p(2*I*(b*x+a))+I)-3/4*I*f^2/b*e*polylog(2,-I*exp(2*I*(b*x+a)))*x^2+3/4*I*f ^2/b^3*e*polylog(2,-I*exp(2*I*(b*x+a)))*a^2+3/4*I*f^2/b*e*polylog(2,I*exp( 2*I*(b*x+a)))*x^2-3/8*I*f^2/b^3*e*polylog(4,I*exp(2*I*(b*x+a)))+1/2*I*f^3/ b^4*a^3*dilog(1+exp(I*(b*x+a))*(-1)^(3/4))+1/2*I*f^3/b^4*a^3*dilog(1-exp(I *(b*x+a))*(-1)^(3/4))-1/4*I*f^3/b*polylog(2,-I*exp(2*I*(b*x+a)))*x^3-1/4*I *f^3/b^4*polylog(2,-I*exp(2*I*(b*x+a)))*a^3+1/2*e^3*ln(1+exp(I*(b*x+a))*(- 1)^(3/4))*x+1/2*e^3*ln(1-exp(I*(b*x+a))*(-1)^(3/4))*x+1/8/f*e^4*ln(-exp(2* I*(b*x+a))+I)+1/8*f^3*ln(1+I*exp(2*I*(b*x+a)))*x^4-3/4*f/b^2*e^2*ln(-I*exp (2*I*(b*x+a))+1)*a^2+3/2*f^2/b^3*a^3*e*ln(1+exp(I*(b*x+a))*(-1)^(3/4))+3/2 *f^2/b^3*a^3*e*ln(1-exp(I*(b*x+a))*(-1)^(3/4))-3/2*f/b^2*a^2*e^2*ln(1+exp( I*(b*x+a))*(-1)^(3/4))-3/2*f/b^2*a^2*e^2*ln(1-exp(I*(b*x+a))*(-1)^(3/4))+3 /4*f/b^2*e^2*ln(1+I*exp(2*I*(b*x+a)))*a^2-1/2*f^2*ln(exp(2*I*(b*x+a))-I...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1809 vs. \(2 (236) = 472\).
Time = 0.33 (sec) , antiderivative size = 1809, normalized size of antiderivative = 5.99 \[ \int (e+f x)^3 \text {arctanh}(\tan (a+b x)) \, dx=\text {Too large to display} \]
-1/32*(3*f^3*polylog(5, (I*tan(b*x + a)^2 + 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) - 3*f^3*polylog(5, (I*tan(b*x + a)^2 - 2*tan(b*x + a) - I)/(ta n(b*x + a)^2 + 1)) + 3*f^3*polylog(5, (-I*tan(b*x + a)^2 + 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)) - 3*f^3*polylog(5, (-I*tan(b*x + a)^2 - 2*tan(b *x + a) + I)/(tan(b*x + a)^2 + 1)) + 4*(-I*b^3*f^3*x^3 - 3*I*b^3*e*f^2*x^2 - 3*I*b^3*e^2*f*x - I*b^3*e^3)*dilog(-((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + 4*(-I*b^3*f^3*x^3 - 3*I*b^3*e*f ^2*x^2 - 3*I*b^3*e^2*f*x - I*b^3*e^3)*dilog(-((I + 1)*tan(b*x + a)^2 - 2*t an(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + 4*(I*b^3*f^3*x^3 + 3*I*b^ 3*e*f^2*x^2 + 3*I*b^3*e^2*f*x + I*b^3*e^3)*dilog(-(-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + 4*(I*b^3*f^3*x^3 + 3*I*b^3*e*f^2*x^2 + 3*I*b^3*e^2*f*x + I*b^3*e^3)*dilog(-(-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(b^4*f^3*x^ 4 + 4*b^4*e*f^2*x^3 + 6*b^4*e^2*f*x^2 + 4*b^4*e^3*x + 4*a*b^3*e^3 - 6*a^2* b^2*e^2*f + 4*a^3*b*e*f^2 - a^4*f^3)*log(((I + 1)*tan(b*x + a)^2 + 2*tan(b *x + a) - I + 1)/(tan(b*x + a)^2 + 1)) - 2*(4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3*b*e*f^2 - a^4*f^3)*log(((I + 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 + 1)) + 2*(4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3* b*e*f^2 - a^4*f^3)*log(((I + 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I - 1) /(tan(b*x + a)^2 + 1)) - 2*(b^4*f^3*x^4 + 4*b^4*e*f^2*x^3 + 6*b^4*e^2*f...
\[ \int (e+f x)^3 \text {arctanh}(\tan (a+b x)) \, dx=\int \left (e + f x\right )^{3} \operatorname {atanh}{\left (\tan {\left (a + b x \right )} \right )}\, dx \]
\[ \int (e+f x)^3 \text {arctanh}(\tan (a+b x)) \, dx=\int { {\left (f x + e\right )}^{3} \operatorname {artanh}\left (\tan \left (b x + a\right )\right ) \,d x } \]
1/16*(f^3*x^4 + 4*e*f^2*x^3 + 6*e^2*f*x^2 + 4*e^3*x)*log(2*cos(2*b*x + 2*a )^2 + 2*sin(2*b*x + 2*a)^2 + 4*sin(2*b*x + 2*a) + 2) - 1/16*(f^3*x^4 + 4*e *f^2*x^3 + 6*e^2*f*x^2 + 4*e^3*x)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 - 4*sin(2*b*x + 2*a) + 2) - integrate(1/2*((b*f^3*x^4 + 4*b*e*f^2* x^3 + 6*b*e^2*f*x^2 + 4*b*e^3*x)*cos(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f^ 3*x^4 + 4*b*e*f^2*x^3 + 6*b*e^2*f*x^2 + 4*b*e^3*x)*sin(4*b*x + 4*a)*sin(2* b*x + 2*a) + (b*f^3*x^4 + 4*b*e*f^2*x^3 + 6*b*e^2*f*x^2 + 4*b*e^3*x)*cos(2 *b*x + 2*a))/(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1), x)
\[ \int (e+f x)^3 \text {arctanh}(\tan (a+b x)) \, dx=\int { {\left (f x + e\right )}^{3} \operatorname {artanh}\left (\tan \left (b x + a\right )\right ) \,d x } \]
Timed out. \[ \int (e+f x)^3 \text {arctanh}(\tan (a+b x)) \, dx=\int \mathrm {atanh}\left (\mathrm {tan}\left (a+b\,x\right )\right )\,{\left (e+f\,x\right )}^3 \,d x \]